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BARSIC [14]
3 years ago
11

Interpret using 12 as a reference point how can you tell that the hands of the clock on the right have moved from the previous p

osition

Physics
1 answer:
zubka84 [21]3 years ago
6 0

Answer:

The answer is "6:45 and 7:15".

Explanation:

In the given attache file two clocks are added, that's explanation defined as follows:

In the left image, it may assume, that the clock only at the right-hand side of a scene is move-in its previous position, which is shown to the left, as it was at 6:45 on the left but in the right image it will display the time that is 7:15.

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A push or pull that causes an object's motion to change is called _____
12345 [234]

Answer:

Force

Explanation:

Force is a pull or push that causes An objects motion to change . When force is applied to an object , it either pushes the object to or pulls it from something thereby changing the motion of the object. A good example is the earth’s gravitational force. This force pulls or attracts a body to the center of the earth . Another example Is experienced when kicking a football from one part of a football field to another. The ball is kicked with force and this causes it to move from one place another .

6 0
3 years ago
When using graphs in science what do x and y axes represent
omeli [17]
X chromosome an the Y chromosome
4 0
3 years ago
At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room t
emmasim [6.3K]

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

R_t = R_o[1 + \alpha(T-T_o)]\\\\

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T  = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

8 0
3 years ago
an elevator suspended by a vertical cable is moving downward but slowing down. the tension in the cable must be:
vladimir1956 [14]

Answer:

F = M a is the vector equation involved

F = T - M g      are the forces acting on the elevator   (scalar equation)

T - M g = M a

T = M (a + g)    remember this a scalar

If a is slowing down then it must have a positive acceleration upwards

Therefor the tension in the cable must be greater than zero

When the tension increases to M g, a has increased to zero

For a to be zero, no acceleration, T = M g

6 0
2 years ago
Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is 180.00 km/h and
snow_lady [41]

Answer:

Explanation:

From the given information:

The coordinate axis is situated in the east and north direction.

So, the north will be the  y-axis and the east will be the x-axis

Similarly, the velocity of the plane in regard to the air in the coordinate system will be v_{P/A} = v( cos \theta \ i + sin \theta \ j)

where:

v_{P/A} = velocity of the plane in regard to the air

v = velocity

θ =  angle of inclination of the plane with respect to the horizontal

replacing v = 180 km/ and θ = 20° in above equation, then:

The velocity of the airplane in the coordinate system as:

v_{P} = v_o( cos \phi \ i + sin \phi \ j)

where;

v_p = velocity of the airplane

v_o = velocity

∅ = angle of inclination with regard to the base axis;

Then; replacing  v_o  = 150 km/h and ∅ = 30°

Therefore, the velocity of the plane in the system is :

v_p = v_A + v_{P/A}

v_A=  v_P  -v_{P/A}   --- (1)

v_A= ( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j

v_A= (-39.24 km/h)i + (13.44 km/h) j

The magnitude is:

v_A= (-39.24 km/h)i + (13.44 km/h) j

|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}

v_A = 41.48 km/h

The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.

The angle of motion is:

tan θ = 39.24/13.44

tan θ = 2.9

θ  = tan ^{-1} (2.9)

θ  =  70.97°

The angle of motion is  70.97° from west of north with a velocity of 41.48 km/h.

5 0
3 years ago
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